Question

Find the antiderivative of (e^x)(sinx)(x) using integration by parts twice.

Answer 1

is this suposed to be

e^x(sin(x))

Answer 2

\[\int\limits_{}^{}e^xsin(x)dx\]

Answer 3

\[\int\limits_{}^{}udv=uv-\int\limits_{}^{}vdu\]

Answer 4

You have to do integration by pars twice

Answer 5

he said that in his problem itself ... anyways what do you want you u and dv to be, bobobo

Answer 6

\[
u = \sin(x)
\\
dv = e^x dx
\]

Answer 7

\[du=\cos(x)\]
\[v=e^x\]

Answer 8

use this information in the formula

Answer 9

\[uv-\int\limits_{}^{}vdu=e^xsin(x)-\int\limits_{}^{}e^xcos(x)dx\]

Answer 10

now once again you have another integral that you must solve by parts again

Answer 11

\[u=\cos(x),dv=e^x\]

du=-sin(x)dx
\[v=e^x\]

Answer 12

using this information use the equation again

\[\int\limits_{}^{}e^xsin(x)dx=e^xsin(x)-(e^xcos(x)-\int\limits_{}^{}-e^xsin(x)dx\]

Answer 13

now first pull the negative out of the integral to get

\[e^xsin(x)-1(e^xcos(x)+\int\limits_{}^{}e^xsin(x)dx)\]

distribute the negative 1 out

\[e^xsin(x)-e^xcos^x-\int\limits_{}^{}e^xsin(x)dx\]

at this point you can see you are back at your original functions however one is negative and you can add the integral to the other side and combine like integrals

\[\int\limits_{}^{}e^xsin(x)dx=e^xsin(x)-e^xcos(x)-\int\limits_{}^{}e^xsin(x)dx\]

\[2\int\limits_{}^{}e^xsin(x)dx=e^xsin(x)-e^xcos(x)\]

divide by two and you can simplify the top by factoring and e^x out

\[\int\limits_{}^{}e^xsin(x)dx=\frac{e^x(\sin(x)-\cos(x))}{2}+c\]

Answer 14

Aha! I see it! Thank you very much. Very detailed and great answer.

Answer 15

yeah basically with these 2 by parts, you want to get the integrations the same so you can add one to another and then divide by 2

Answer 16

One question. When you put ∫e^xcos(x)dx into the equation, why did it stay as e^xcos(x)? Why not -e^xsinx?

Answer 17

where?

Answer 18

So, I will talk in terms of f' and g'. If youve got the integral of e^xcosx, you will let e^x=f and cosx=g' right? So it should be -e^xsinx right?

Answer 19

yes that would be the second one but cos(x) i made f and e^x = g'

Answer 20

thats how when i did the second integration i got

-e^x(sin(x))

Answer 21

the first time i did it i made sin(x)=f and e^x=g'

derivative of sin(x) = cos(x)

Answer 22

Okay, I see it. Thanks :)

Answer 23

no problem =]